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Angelina_Jolie [31]
3 years ago
9

What is the medium through which the wave is moving?

Physics
2 answers:
irakobra [83]3 years ago
5 0

Rope or one could consider it to be C to B or A to D

emmasim [6.3K]3 years ago
4 0

Answer:

STRING held by the man

Explanation:

Wave is defined as a disturbance that travels through a MEDIUM and transfer energy from one point to another without causing any permanent displacement of the medium itself.

The medium through which wave travels varies e.g water, string, air etc

According to the diagram, wave travels through the string held by the man. This string is referred to as the MEDIUM through which the wave moves. The wave generated produces both crest (D) and trough (C) when displaced from its initial position (A)

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Which two terms indicate the same area of the body? *
Kruka [31]

Answer:

For example, the toes are anterior to the heel, and the popliteus is posterior to the patella. Superior and inferior, which describe a position above (superior) or below (inferior) another part of the body. For example, the orbits are superior to the oris, and the pelvis is inferior to the abdomen.

Explanation:

4 0
3 years ago
Does kinetic energy stay the same at all heights? pls help I have 12 minutes to finish the project and the teacher won't help me
Alex Ar [27]

Answer:

Kinetic energy does not stay the same at all heights

Explanation:

Well as the height and wind increase so does the kinetic energy it's like when you fall as you are about to hit the floor you speed increases

HOPE THIS HELPS YA :)

7 0
3 years ago
9.58 A spring of equilibrium length L1 and spring constant k1 hangs from the ceiling. Mass m1 is suspended from its lower end. T
andrey2020 [161]

Answer:

The distance of m2 from the ceiling is L1 +L2 + m1g/k1 + m2g/k1 + m2g/k2.

See attachment below for full solution

Explanation:

This is so because the the attached mass m1 on the spring causes the first spring to stretch by a distance of m1g/k1 (hookes law). This plus the equilibrium lengtb of the spring gives the position of the mass m1 from the ceiling. The second mass mass m2 causes both springs 1 and 2 to stretch by an amout proportional to its weight just like above. The respective stretchings are m2g/k1 for spring 1 and m2g/k2 for spring 2. These plus the position of m1 and the equilibrium length of spring 2 L2 gives the distance of L2 from the ceiling.

4 0
3 years ago
A pizza delivery driver must make three stops on her route. She will first leave the restaurant and travel 4 km due north to the
topjm [15]
<h2>5.3 km</h2>

Explanation:

       This question involves continuous displacement in various directions. When it becomes difficult to imagine, vector analysis becomes handy.

       Let us denote each of the individual displacements by a vector. Consider the unit vectors \vec{i}\textrm{ and }\vec{j} as the unit vectors in the direction of East and North respectively.

       By simple calculations, we can derive the unit vectors \vec{j},\frac{-\vec{i}-\vec{j}}{2}\textrm{ and }\frac{-\frac{1}{2}\vec{i}+\frac{\sqrt{3}}{2}\vec{j}}{2} in the directions North, 45^{o} South of West and 60^{o} North of West respectively.

       So Total displacement vector = Sum of individual displacement vectors.

       Displacement vector = 4(\vec{j})+6(\frac{-\vec{i}-\vec{j}}{2})+5(\frac{-\frac{1}{2}\vec{i}+\frac{\sqrt{3}}{2}\vec{j}}{2})=-4.25\vec{i}+3.165\vec{j}

       Magnitude of Displacement = |-4.25\vec{i}+3.165\vec{j}|=5.3km

∴ Total displacement = 5.3km

4 0
3 years ago
To practice problem-solving strategy 22.1: gauss's law. an infinite cylindrical rod has a uniform volume charge density ρ (where
BabaBlast [244]

Let say the point is inside the cylinder

then as per Gauss' law we have

\int E.dA = \frac{q}{\epcilon_0}

here q = charge inside the gaussian surface.

Now if our point is inside the cylinder then we can say that gaussian surface has charge less than total charge.

we will calculate the charge first which is given as

q = \int \rho dV

q = \rho * \pi r^2 *L

now using the equation of Gauss law we will have

\int E.dA = \frac{\rho * \pi r^2* L}{\epcilon_0}

E. 2\pi r L = \frac{\rho * \pi r^2* L}{\epcilon_0}

now we will have

E = \frac{\rho r}{2 \epcilon_0}

Now if we have a situation that the point lies outside the cylinder

we will calculate the charge first which is given as it is now the total charge of the cylinder

q = \int \rho dV

q = \rho * \pi r_0^2 *L

now using the equation of Gauss law we will have

\int E.dA = \frac{\rho * \pi r_0^2* L}{\epcilon_0}

E. 2\pi r L = \frac{\rho * \pi r_0^2* L}{\epcilon_0}

now we will have

E = \frac{\rho r_0^2}{2 \epcilon_0 r}


7 0
3 years ago
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